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          状态反馈思考与仿真
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            <div class="post-description">学过状态反馈，但从未细致的理解过反馈于何处，系统在加入状态反馈的前后怎样的变化。最近复习现代控制理论，偶然间理解了更多，在此做出记录。主要记录了状态反馈、系统的可控性，最后用仿真实验联系上述的内容加强理解。</div>

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        <h1 id="状态反馈">状态反馈</h1>
<p>系统的状态方程如下： <span class="math display">\[
\begin{equation}
    \begin{aligned}
         \dot{x} &amp;= Ax+Bu\\
          y &amp;= Cx
    \end{aligned}
\end{equation}
\]</span> 其方框图如下：</p>
<p><img src="image-20210910214357380.png" alt="image-20210910214357380" style="zoom:25%;" /></p>
<p>状态反馈指把状态变量 <span class="math inline">\(x\)</span> 与输入 <span class="math inline">\(u\)</span> 相减，获得新的系统和其框图如下： <span class="math display">\[
\begin{equation}
    \begin{aligned}
         \dot{x} &amp;= (A-BK)x+Bu\\
          y &amp;= Cx
    \end{aligned}
\end{equation}
\]</span> <img src="image-20210910214701286.png" alt="image-20210910214701286" style="zoom:25%;" /></p>
<p>使用状态反馈后，系统不再是原来的系统，变成了新的系统，新系统的极点由 <span class="math inline">\(K\)</span>​ 决定。特别的，<strong>如果原系统能控，则状态反馈可以把系统的极点调整到任何一点。</strong></p>
<p>此外新系统的输入不再是 <span class="math inline">\(u\)</span> 而是 <span class="math inline">\(v\)</span>，可以理解为状态反馈使得系统变为闭环。这里的方框图还很概括，无法显示状态反馈的每一处细节，在仿真中搭建的图会更为具体。</p>
<h1 id="系统的能控性">系统的能控性</h1>
<p>系统的可控性是指：系统在<span class="math inline">\(t_0\)</span>​​时刻的状态为<span class="math inline">\(x_0\)</span>​​，<span class="math inline">\(t_1\)</span>​​时刻的状态为<span class="math inline">\(x_1\)</span>；是否存在一组有限大的输入 <span class="math inline">\(u\)</span>，使得系统在 <span class="math inline">\(t_0 - t_1\)</span> 时间内，从 <span class="math inline">\(t_0\)</span> 变为 <span class="math inline">\(t_1\)</span>。</p>
<p>可用能控性判别矩阵<span class="math inline">\(C_o = [B~AB~A^2B ... A^{n-1}B]\)</span>​，来判别系统是否能控，如果能控，则矩阵满秩。​​</p>
<p>注意：系统的能控性是指系统的状态变量可在有限时间内到达任意一点，而不是说系统稳定，这是两个概念。</p>
<h1 id="仿真例子">仿真例子</h1>
<h2 id="例子1">例子1</h2>
<p>考虑一个小车模型如下：</p>
<p><img src="image-20210911113226022.png" alt="image-20210911113226022" style="zoom: 15%;" /></p>
<p>小车质量为<span class="math inline">\(m_1\)</span>​，位移为<span class="math inline">\(x_1\)</span>​，外部输入力为<span class="math inline">\(u = F(t)\)</span>​；期望通过<span class="math inline">\(u\)</span>​控制小车位移<span class="math inline">\(x_1\)</span>​和速度<span class="math inline">\(\dot{x}_1\)</span>​。</p>
<p>首先对系统建模，使用牛顿第二定律： <span class="math display">\[
u = m \ddot{x}
\]</span> 设<span class="math inline">\(x_1 = x,x_2 = \dot{x_1}\)</span>，得到系统的状态方程如下： <span class="math display">\[
\dot{x} = 
\begin{bmatrix}
0 &amp; 1 \\
0 &amp; 0 
\end{bmatrix}
x +
\begin{bmatrix}
0 \\
1 
\end{bmatrix}u
\]</span> 计算能控性判别矩阵为： <span class="math display">\[
C_o = \begin{bmatrix}B &amp; AB \end{bmatrix}=\begin{bmatrix}0 &amp; 1 \\1 &amp; 0 \end{bmatrix} 
\]</span> 满秩；所以系统可控。但系统的特征值为<span class="math inline">\(\lambda = 0\)</span>​​​​，系统不稳定。​在simulink中搭建模型如下：(设<span class="math inline">\(m=1\)</span>​​)</p>
<figure>
<img src="image-20210911120339190.png" alt="" /><figcaption>image-20210911120339190</figcaption>
</figure>
<p>实际上，要搭建这样的模型，对照微分方程更容易理解： <span class="math display">\[
\left\{\begin{array}{l}
\dot{x}_{1}=x_{2} \\
u=m \ddot{x_{1}}=\dot{x}_{2}
\end{array}\right.
\]</span> 这里的<span class="math inline">\(\dot{x}_2\)</span>就是外部输入，当无外外部输入时，系统的响应如下：</p>
<p><img src="image-20210911121018454.png" alt="image-20210911121018454" style="zoom: 67%;" /></p>
<p>是一直为0的直线，从物理解释，这很好理解，当没有外力作用时，小车保持静止，位移和速度都为0；这也是原系统的效果。取消注释阶跃信号，系统响应如下：</p>
<p><img src="image-20210911121327457.png" alt="image-20210911121327457" style="zoom:50%;" /></p>
<p>两条曲线发散，从物理解释，这很好理解，当有一个力<span class="math inline">\(F\)</span>一直作用于小车，小车会做匀加速直线运动；速度是关于时间的一次函数，位移是时间的二次函数。</p>
<h3 id="加入状态反馈">加入状态反馈</h3>
<p>这个系统不能达到我们的目的——控制小车的速度和位移。其中一部分原因是原系统实际上是一个开环系统，拉动小车的力是随机的。现在我们加入状态反馈使得系统闭环。</p>
<p>这是一个二阶系统，且能控，现在把系统的极点配置到<span class="math inline">\(-1,-2\)</span>​。系统的期望多项式为： <span class="math display">\[
(s+1)(s+2) = s^2 + 3s+2
\]</span> 设<span class="math inline">\(K=[k_1 ~ k_2]\)</span>，于是<span class="math inline">\(|SI-(A-BK)| = s^2 + k_2s+k_1\)</span>，对比系数，有<span class="math inline">\(k_1 = 2,k_2=3\)</span>。于是仿真图型变为下图：</p>
<figure>
<img src="image-20210911162525833.png" alt="" /><figcaption>image-20210911162525833</figcaption>
</figure>
<p>从仿真图来看，速度和位移在倍乘后直接变为u，若<span class="math inline">\(v=0\)</span>，则<span class="math inline">\(u = -Kx\)</span>​。设置速度<span class="math inline">\(x_2\)</span>的初始值为1，仿真结果如下：</p>
<p><img src="image-20210911162756991.png" alt="image-20210911162756991" style="zoom: 67%;" /></p>
<p>可见加速度、位移、速度最后都变为0，控制效果还是不错的。</p>
<h2 id="例子2">例子2</h2>
<p>考虑两个小车模型如下：</p>
<p><img src="image-20210911163815643.png" alt="image-20210911163815643" style="zoom: 15%;" /></p>
<p>期望通过<span class="math inline">\(u\)</span>控制小车位移<span class="math inline">\(x_1，x_2\)</span>和速度<span class="math inline">\(\dot{x}_1，\dot{x}_2\)</span>​。首先进行建模 <span class="math display">\[
\left\{\begin{array}{l}
m_{2} \ddot{x}_{2}=k\left(x_{1}-x_{2}\right) \\
m_{1} \ddot{x}_{1}=u-k\left(x_{1}-x_{2}\right)
\end{array}\right.
\]</span> 设<span class="math inline">\(m_1=m_2=1kg,k=100N/m\)</span>；设<span class="math inline">\(z_1 = x_1,z_2 = \dot{x_1},z_3 = x_2,z_4 = \dot{x_2}\)</span>​，得系统状态方程： <span class="math display">\[
\dot{z} = 
\begin{bmatrix}
0 &amp; 1 &amp; 0 &amp; 0\\
-100 &amp; 0 &amp; 100 &amp; 0\\
0 &amp; 0 &amp; 0 &amp; 1\\
100 &amp; 0 &amp; -100 &amp; 0\\
\end{bmatrix}
z +
\begin{bmatrix}
0 \\
1 \\
0 \\
0
\end{bmatrix}u
\]</span> 计算可知，这个系统是能控的。其特征值为<span class="math inline">\(\lambda_{1,2}=0,\lambda_{3,4}=14.1421i\)</span>​，很明显​系统不可控。仿真图如下：</p>
<p><img src="image-20210912110756048.png" /></p>
<p>当设置<span class="math inline">\(z_1\)</span>的初始值为1时，得到系统响应如下，因为系统的极点都在虚轴上，所以会等幅震荡。</p>
<p><img src="image-20210912110831227.png" alt="image-20210912110831227" style="zoom:50%;" /></p>
<h3 id="加入状态反馈-1">加入状态反馈</h3>
<blockquote>
<p>这是一个四阶系统，可以设计两个远离虚轴的极点，和两个<strong>靠近虚轴的闭环主导极点</strong>，系统的瞬态性能指标主要由主导极点决定。 这样就可以<strong>把一个高阶系统简化成一个二阶系统</strong>。</p>
</blockquote>
<p>二阶系统的一般形式为：<span class="math inline">\(\frac{C(s)}{R(s)}=\frac{\omega_{n}^{2}}{s^{2}+2 \zeta \omega_{n} s+\omega_{n}^{2}}\)</span></p>
<p>希望调节时间<span class="math inline">\(t_s=2s\)</span>，超调量为<span class="math inline">\(\delta=0.5\)</span>，套用公式<span class="math inline">\(t_{\mathrm{s}} \approx \frac{4}{\zeta \omega_{\mathrm{n}}} ，\delta \%=\mathrm{e}^{-\frac{5}{\sqrt{1-5^{2}}}} \times 100 \%\)</span>。可得两个主导极点为<span class="math inline">\(\lambda_{1,2}=-2 \pm 2 \sqrt{3}i\)</span>。设计两个远离虚轴极点为-10.期望多项式为： <span class="math display">\[
s^{4}+24 s^{3}+196 s^{2}+720 s+1600=0
\]</span> 设<span class="math inline">\(K=[k_1~k_2~k_3~k_4]\)</span>，所以 <span class="math display">\[
|SI-(A-BK)| = s^4 + k_2s^3+(200+k_1)s^2+(100k_2+100k_4)s+100k_1+100k_3
\]</span> 对照有：<span class="math inline">\(k_1 =-4,k_2=24,k_3=20,k_4=-16.8\)</span>​​.仿真图如下：</p>
<p><img src="image-20210912111540351.png" /></p>
<p>对于一个四阶系统： <span class="math display">\[
\dot{x} =
\begin{bmatrix}
a_{11} &amp; a_{12} &amp; a_{13} &amp; a_{14}\\
a_{21} &amp; a_{22} &amp; a_{23} &amp; a_{24}\\
a_{31} &amp; a_{32} &amp; a_{33} &amp; a_{34}\\
a_{41} &amp; a_{42} &amp; a_{43} &amp; a_{44}\\
\end{bmatrix}
x +
\begin{bmatrix}
b_1\\
b_2 \\
b_3\\
b_4\\
\end{bmatrix}
u
\]</span> 加入状反馈矩阵<span class="math inline">\(K=[k_1~k_2~k_3~k_4]\)</span>后变为： <span class="math display">\[
\dot{x} =
\begin{bmatrix}
a_{11}-b_1k_1 &amp; a_{12}-b_1k_2 &amp; a_{13}-b_1k_3 &amp; a_{14}-b_1k_4\\
a_{21}-b_2k_1 &amp; a_{22}-b_2k_2 &amp; a_{23}-b_2k_3 &amp; a_{24}-b_2k_4\\
a_{31}-b_3k_1 &amp; a_{32}-b_3k_2 &amp; a_{33}-b_3k_3 &amp; a_{34}-b_3k_4\\
a_{41}-b_4k_1 &amp; a_{42}-b_4k_2 &amp; a_{43}-b_4k_3 &amp; a_{44}-b_4k_4\\
\end{bmatrix}
x +
\begin{bmatrix}
b_1\\
b_2 \\
b_3\\
b_4\\
\end{bmatrix}
u
\]</span> 也就是说，每个<span class="math inline">\(\dot{x}\)</span>​都会接收其余状态变量反馈的值。在本例子中，只有<span class="math inline">\(b_2 \neq0\)</span>​，所以<strong>只有<span class="math inline">\(\dot{x}_2\)</span>​会接收状态反馈</strong>。当设置<span class="math inline">\(z_1\)</span>​的初始值为1时，得到系统响应如下，可见控制效果达到预期：</p>
<p><img src="image-20210912111852100.png" alt="image-20210912111852100" style="zoom:50%;" /></p>
<p>如果在<code>add2</code>处加入一个阶跃信号，相当于输入<span class="math inline">\(v\)</span>一个阶跃信号，系统的响应如下，位置改变但速度仍然为0；但我暂时还未明白这代表了什么意义。</p>
<p><img src="image-20210912161727114.png" alt="image-20210912161727114" style="zoom: 50%;" /></p>
<h1 id="小结">小结</h1>
<p>这篇博文记录了我近期复习现代控制理论的思考；在搭建仿真模型中，对状态反馈有了更深入的理解。所谓状态反馈，就是把系统的状态进行数乘，作为输入信号输入到系统中。可以说，状态反馈代替了原本的输入信号。</p>
<p>这里，系统的输入矩阵<span class="math inline">\(B\)</span>显得格外重要，首先<span class="math inline">\(B\)</span>会影响系统本来的输入对状态变量的效果。比如例2中，<span class="math inline">\(B=[0~1~0~0]^T\)</span>，这就使得输入<span class="math inline">\(u\)</span>只能对<span class="math inline">\(\dot{x}_2\)</span>起作用；其次在使用状态反馈时，反馈回的状态变量也只能对<span class="math inline">\(\dot{x}_2\)</span>起作用。</p>
<p>但是仍然留有疑问，上面的例子中都是定值系统，在系统有扰动加入或系统初始状态不为0时，最终系统都是恢复0状态。如何构造随动系统呢？这点我还没有想明白，需要继续思考。</p>

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